UVA 11127 - Triple-Free Binary Strings

Problem J
Triple-Free Binary Strings 
Input: Standard Input

Output: Standard Output

 

A binary string consists of ones and zeros. Given a binary string T, if  there is no binary string S such that SSS (concatenate three copies of S together) is a substring of T, we say T is triple-free.

A pattern consists of ones, zeros and asterisks, where an asterisk(*) can be replaced by either one or zero. For example, the pattern 0**1 contains strings 0001, 0011, 0101, 0111, but not 1001 or 0000.

Given a pattern P, how many triple-free binary strings does it contain?

InputEach line of the input represents a test case, which contains the length of pattern, n(0<n<31), and the pattern P. There can be maximum 35 test cases. 

The input terminates when n=0.

OutputFor each test case, print the case number and the answer, shown below. 

 

Sample Input                       Output for Sample Input

4 0**1

5 *****

10 **01**01**

0

Case 1: 2

Case 2: 16

Case 3: 9

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Problemsetter: Rujia Liu
Special Thanks: Shahriar Manzoor

hash 暴力搜一下就过了....不多说了

#include <cstdio>#include <iostream>#include <cstring>#include <string.h>#include <algorithm>#include <set>#include <vector>using namespace std;const int maxn = 30 + 5;char S[maxn];int n;char A[maxn];int H[maxn];vector<int> X[16];#define rep(i,a,b) for(int i = (a); i < (b); ++i)#define rrep(i,b,a) for(int i = (b); i >= (a); --i)#define clr(a,x) memset(a,x,sizeof(a))int Hash(int l,int r){    return H[r] - (H[l-1] << (r-l+1));}bool Ok(int cur){    if (S[cur] != '*' && A[cur] != S[cur]) return false;    for(int len = 1; cur - len * 3 >= 0; ++len) {        int h = Hash(cur-3*len+1,cur);        if (binary_search(X[len].begin(),X[len].end(),h)) return false;    }    return true;}int dfs(int cur){    if (cur == n+1) return 1;    A[cur] = '1';    H[cur] = (H[cur-1]<<1) + 1;    int sum = 0;    if (Ok(cur)) {        sum += dfs(cur+1);    }    A[cur] = '0';    H[cur] = H[cur-1] << 1;    if (Ok(cur)) {        sum += dfs(cur+1);    }    return sum;}void solve(){    clr(H,0);    int ans = dfs(1);    printf("%d\n",ans);}int main(){    rep(i,1,11) {        rep(j,0,1<<i) {            int x = j + (j << i) + (j << (2*i));            X[i].push_back(x);        }        sort(X[i].begin(),X[i].end());    }    int cas = 0;    while (scanf("%d",&n),n) {        scanf("%s",S+1);        ++cas;        printf("Case %d: ",cas);        solve();    }    return 0;}